2475. Number of Unequal Triplets in Array
Explanation
To solve this problem, we can iterate through all possible triplets (i, j, k) and check if they meet the conditions. We need to make sure that the elements at indices i, j, and k are pairwise distinct. We can achieve this by using three nested loops. If the conditions are met, we increment a counter to keep track of the number of valid triplets. Finally, we return the count of valid triplets.
- Time Complexity: O(n^3) where n is the length of the input array nums.
- Space Complexity: O(1) as we are using constant extra space.
class Solution {
public int countGoodTriplets(int[] nums) {
int n = nums.length;
int count = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
for (int k = j + 1; k < n; k++) {
if (nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k]) {
count++;
}
}
}
}
return count;
}
}Code Editor (Testing phase)
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